Question

Estimate the pI for the peptide Arg-Met-His-Val-Glu. Please show work and explanation.

Answer 1

At pH = pK_a, half of the acid is deprotonated.

At pH = pI, the net charge on the molecule = 0. At this pH, the peptide will not migrate in an electric field. Then, that pH is called the isoelectric point.

At pH = 0, the net charge of the molecule = +3. At pH between 2.19 and 4.25, the carboxylic acid group at the C-terminus will get deprotonated and then the net charge on the molecule = (+3-1) = +2

At pH between 4.25 and 6.00, the side chain carboxylic acid group of the glutamic acid residue will be deprotonated resulting in the net charge of (+3-2) = +1 on the molecule.

At pH between 6.00 and 9.04, the imidazole moiety of the histidine residue will be deprotonated resulting in the net charge of (+2-2) = 0 on the molecule.

Hence, the isoelectric point pI = (6.00 + 9.04)/2 = 7.52

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